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# Nesbitt's Inequality: Alternative Proof

I recently came across the well-known Nesbitt's inequality while reading about famous mathematical relations. If you visit the Wikipedia page for this inequality, there are six different proofs for it. I then proceeded to prove it in a different way, that is, the method I used was not articulated on the Wikipedia page. The proof I wrote is below, I thought Brilliant users would find it somewhat interesting (if you find any errors in it, please point them out as I'm trying to improve my proof writing skills):

Nesbitt's Inequality:

$$\forall\ a,b,c \in \mathbb{R}^{+} : \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{3}{2}$$

Proof:

Let $$f(a,b,c)= \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b}$$. We will show that $$f(a,b,c)$$ has a minimum when $$a=b=c$$ and that all other extrema imply that at least one of $$a,b,c \notin \mathbb{R}^{+}$$.

All maxima and minima occur when $$\nabla f(a,b,c)=\vec{0}$$. Hence we find the partial derivatives of $$f(a,b,c)$$:

$$\frac{\partial f}{\partial a} = \frac{1}{b+c} -\frac{b}{(a+c)^2}-\frac{c}{(a+b)^2}$$

$$\frac{\partial f}{\partial b}= -\frac{a}{(b+c)^2}+\frac{1}{a+c}-\frac{c}{(a+b)^2}$$

$$\frac{\partial f}{\partial c}= -\frac{a}{(b+c)^2}-\frac{b}{(a+c)^2}+\frac{1}{a+b}$$

We now set $$\nabla f(a,b,c)=\vec{0}$$:

$$0=\frac{1}{b+c} -\frac{b}{(a+c)^2}-\frac{c}{(a+b)^2}$$

$$0=-\frac{a}{(b+c)^2}+\frac{1}{a+c}-\frac{c}{(a+b)^2}$$

$$0=-\frac{a}{(b+c)^2}-\frac{b}{(a+c)^2}+\frac{1}{a+b}$$

Solving this non-linear system:

$$\frac{1}{a+c}+\frac{b}{(a+c)^2} = \frac{1}{a+b} + \frac{c}{(a+b)^2}$$ $$\Rightarrow$$ $$(a+c)^2=(a+b)^2$$ $$\Rightarrow$$ $$c=-a \pm (a+b)$$ $$\Rightarrow$$ $$c=b$$ or $$c=-2a-b$$

$$c=-2a-b$$ implies $$c \notin \mathbb{R}^{+}$$, since we assume $$a,b \in \mathbb{R}^{+}$$. Hence we choose $$c=b$$.

$$\frac{1}{b+c}-\frac{b}{(a+c)^2}=\frac{1}{a+c}-\frac{a}{(b+c)^2}$$ $$\Rightarrow$$ $$\frac{a+b+c}{(b+c)^2}=\frac{a+b+c}{(a+c)^2}$$ $$\Rightarrow$$ $$a=-b \pm 2b$$ $$\Rightarrow$$ $$a=-3b$$ or $$a=b$$.

$$a=-3b$$ implies $$a \notin \mathbb{R}^{+}$$ since we assume $$b \in \mathbb{R}^{+}$$. Then $$a=b$$ $$\Rightarrow$$ $$a=b=c$$.

Since all other solutions of this system require that at least one of $$a,b,c$$ is negative, the only solution such that $$a,b,c \in \mathbb{R}^{+}$$ is $$a=b=c$$, which actually represents infinitely many solutions. Note that $$f(a,b,c)=\frac{3}{2}$$ if $$a=b=c$$. Also note that this must be a minimum, because if we let $$a=1,b=2,c=3$$, then $$f(a,b,c)= 1.7 \geq \frac{3}{2}$$. Then it follows that $$\forall\ a,b,c \in \mathbb{R}^{+} : \frac{a}{b+c} + \frac{b}{a+c} + \frac{c}{a+b} \geq \frac{3}{2}$$, which was to be proved.

QED

Note by Ethan Robinett
3 years, 5 months ago

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I was actually the one who wrote the sixth proof on wikipedia :P

- 3 years, 5 months ago

Lol that's actually pretty cool

- 3 years, 5 months ago

There can also be a seventh proof using convex functions but I am too lazy to write it. : |

- 3 years, 5 months ago